How many ten-digit binary numbers have exactly three 1’s?
What I have find is that some common patterns are they have three 1s and seven 0s. I have an idea by common sense. But, how to prove it mathematically.
This is what I got
C(7,3) = 35
Not sure if go for permutation or combination
$\frac{10!}{7!3!}=120$
Sorry I am on mobile so I cannot write much. But there $3$ 1's and $7$ 0's. So by the permutation formula.
Permutations:$\frac{n!}{r_{1}!r_{2}!...r_{n}!}$
There are 10 ways to put the first 1 in a 10 bit binairy number. Pick any position for the first 1. Then there are 9 slots left for the second 1. Again pick any position... Now there are 8 slots left for the a 1.
So I think the answer could be 10x9x8=720 combinations of binairy numbers containing exactly three ones.
